The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life.

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The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life.

Which brand will you prefer based on this calculation

Car                      Brand   1                     Brand 2

1                          36,925                        34,318

2                         45,300                        42,280

3                         36,240                        35.500

4                         32,100                        31,950

5                         37,210                        38,015

6                         48,360                        47,800

7                         38,200                        37,810

8                         33,500                        33,215

Answer:

the  99% confidence interval  on the difference in the mean life. =

[tex]- 727.46 < \mu_d < 2464.21[/tex]

Step-by-step explanation:

Car       Brand   1             Brand 2        d = Brand (1) - Brand (2)

1           36,925                34,318              2607

2          45,300                42,280             3020

3          36,240                 35.500            740

4          32,100                  31,950             150

5          37,210                  38,015           - 805

6          48,360                 47,800            560

7          38,200                 37,810             390

8          33,500                 33,215             285

The mean of the above value of d is calculated as:

Mean (d') = [tex]\frac {\sum d}{n}[/tex]

= [tex]\frac{2607 + 3020+740+150+(-805)+ 560+390+285}{8}[/tex]

= [tex]\frac{6947}{8}[/tex]

= 868.375

Standard deviation

[tex](S_d) = \sqrt{\frac{(2607-868.375)^2+(3020-868.375)^2+(740-868.375)^2+...+(285-868.375)^2}{8-1}}[/tex]

[tex](S_d) = \sqrt{1664185.41}[/tex]

[tex](S_d) = 1290.03[/tex]

The degree of freedom is calculated as:

[tex]df = n-1 \\df= 8-1\\df = 7[/tex]

Now, from the t - tabulated value;

t-critical value at 99% confidence level  at 99% confidence level with 7 degree of freedom = [tex]\pm \ 3.499[/tex]

In order to determine the 99% confidence interval on the difference in mean life ; we have:

[tex]99[/tex]% C.I = [tex]d' + I_{critical} ( {\frac{S_d}{\sqrt{n}} })[/tex]

[tex]99[/tex]% C.I = [tex]868.375 \pm3.499 ( {\frac{1290.03}{\sqrt{8}} })[/tex]

[tex]99[/tex]% C.I = [tex]868.375 \pm 1595.835[/tex]

[tex]99[/tex]% C.I = ( -727.46 ; OR; 2464.21 )

Thus; the  99% confidence interval  on the difference in the mean life. =

[tex]- 727.46 < \mu_d < 2464.21[/tex]

CONCLUSION:

We therefore conclude that the 99% confidence interval contains null values, as such there is no significant difference between the two brands of tire.