Chapter 12.2, Problem 42E

(a)

To determine

To Calculate: and state the least regression line.

Answer to Problem 42E

  $\log y=-0.2690+1.2609\log x$

Explanation of Solution

Given:

Tread depth (1/32 inch)	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
2	15.2
1	18.3

Calculation:

Taking the log of the given data

Tread depth (1/32 inch)	Breaking distance (car lengths)	log(Tread depth)	log(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
7	10.8	0.84509804	1.03342376
6	11.2	0.77815125	1.04921802
5	11.8	0.698970004	1.07188201
4	12.4	0.602059991	1.09342169
3	13.6	0.477121255	1.13353891
2	15.2	0.301029996	1.18184359
1	18.3	0	1.26245109

Using Ti83/84 calculator

Step 1: press on STAT,

Step 2: choose 1: EDIT

Step 3: type the data of logarithmic of Tread depth in list L1 and in L2 logarithmic of Breaking Distance in L2.

Step 4: again press on STAT, chose CALC and chose LinReg(a+bx)

Final result

  $\begin{array}{l}y=a+bx\\ a=-0.2690\\ b=1.2609\end{array}$

Putting the value in a and b

  $y=-0.2690+1.2609x$

  $x$ is representing the logarithm of tread depth and y is representing the logarithm of Breaking distance

  $\log y=-0.2690+1.2609\log x$

  $x$ Represents the tread depth and y represent the breaking distance.

(b)

To determine

To Calculate: and explain the residual for the trial if the trail when the tread depth is 3/32 inch.

Answer to Problem 42E

2.1508 car length

Explanation of Solution

Given:

Tread depth (1/32 inch)	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
2	15.2
1	18.3

Calculation:

From the part (b)

  $\log y=-0.2690+1.2609\log x$

Putting the value x by 3

  $\begin{array}{l}\log y=-0.2690+1.2609\log x\\ \log y=-0.2690+1.2609\log 3\\ \log y=0.3326\end{array}$

Taking the exponential with base 10

  $\begin{array}{c}y={10}^{\log y}={10}^{0.33263}\\ =2.1508\end{array}$

Therefore the expected breaking distance is 2.1508 car length.